Hybridisation of Carbon — sp, sp² and sp³ with Shapes
Hybridisation feels abstract until you realise it is really a counting exercise. Once you can look at any carbon in any structure and name its hybridisation in three seconds, a whole set of questions opens up: bond angles, molecular shape, bond lengths, relative acidity, and why graphite conducts electricity while diamond does not.
Why carbon needs hybridisation at all
The ground-state electron configuration of carbon is 1s² 2s² 2p². Count the unpaired electrons: only two, both in 2p orbitals. On that basis carbon should form two bonds at about 90°. But methane exists, it has four identical C–H bonds, and all its angles are 109.5°.
The fix has two parts. One 2s electron is promoted to the empty 2p orbital, giving 2s¹ 2p³ — four unpaired electrons, so four bonds are possible. Then those four orbitals (one s and three p) are mixed to produce four new, identical orbitals pointing as far apart as possible. That mixing is hybridisation — a model used to rationalise geometries that were measured experimentally, not a physical event happening in sequence inside the atom.
The rule that generates all three types
1 s + 3 p → four sp³ orbitals
1 s + 2 p → three sp² orbitals (one p orbital left unhybridised)
1 s + 1 p → two sp orbitals (two p orbitals left unhybridised)
The leftover unhybridised p orbitals are not wasted. Each one forms a pi (π) bond by sideways overlap. This is why sp² carbon has exactly one double bond and sp carbon has two π bonds — either two double bonds, or one triple bond.
The three states side by side
| Property | sp³ | sp² | sp |
|---|---|---|---|
| Orbitals mixed | 1 s + 3 p | 1 s + 2 p | 1 s + 1 p |
| Hybrid orbitals formed | 4 | 3 | 2 |
| Unhybridised p orbitals | 0 | 1 | 2 |
| Sigma bonds on that carbon | 4 | 3 | 2 |
| Pi bonds on that carbon | 0 | 1 | 2 |
| Geometry | Tetrahedral | Trigonal planar | Linear |
| Bond angle | 109.5° | 120° | 180° |
| s character | 25% | 33.3% | 50% |
| Simplest example | CH₄, ethane | Ethene, benzene | Ethyne, CO₂ |
The shapes in words: sp³ is three-dimensional — the carbon sits at the centre of a tetrahedron and its four bonds point to the four corners, none of them in one plane. sp² is flat — three bonds spread out in a single plane like the letter Y, with the leftover p orbital standing perpendicular through that plane. sp is a straight line — two bonds in exactly opposite directions, with the two leftover p orbitals at right angles to that line and to each other.
How to identify hybridisation in three seconds
Do not try to picture orbitals in an exam. Count instead. For the carbon in question:
Steric number 4 → sp³ · 3 → sp² · 2 → sp
One working note makes this foolproof: pi bonds are never counted. A double bond is one sigma plus one pi, so it contributes 1 to the count, not 2; a triple bond is one sigma plus two pi, so it also contributes 1.
A quicker shortcut for neutral carbon follows from the same rule: no multiple bond means sp³; one double bond means sp²; one triple bond, or two double bonds on the same carbon, means sp.
Worked example 1 — the three simplest hydrocarbons
Ethane, CH₃–CH₃. Each carbon has 3 sigma bonds to H plus 1 sigma bond to the other carbon, and no lone pairs. Steric number = 4 + 0 = 4 → sp³, tetrahedral, H–C–H angle about 109.5°.
Ethene, CH₂=CH₂. Each carbon has 2 sigma bonds to H plus 1 sigma to the other carbon (the second half of the double bond is a pi bond and is not counted). Steric number = 3 → sp², trigonal planar, angles about 120°. The molecule is flat, and the two unhybridised p orbitals overlap sideways to give the π bond — which locks the ends together and restricts rotation about C=C, the origin of cis–trans isomerism.
Ethyne, CH≡CH. Each carbon has 1 sigma bond to H plus 1 sigma to the other carbon (the other two components of the triple bond are π). Steric number = 2 → sp, linear, H–C≡C–H angle 180°. All four atoms lie on one straight line.
Worked example 2 — harder cases students get wrong
| Species | Carbon considered | Sigma + lone pairs | Hybridisation | Shape at that carbon |
|---|---|---|---|---|
| Carbon dioxide, O=C=O | the carbon | 2 + 0 = 2 | sp | Linear, 180° |
| Hydrogen cyanide, H–C≡N | the carbon | 2 + 0 = 2 | sp | Linear, 180° |
| Benzene, C₆H₆ | every ring carbon | 3 + 0 = 3 | sp² | Trigonal planar, 120° |
| Carbonate ion, CO₃²⁻ | the central carbon | 3 + 0 = 3 | sp² | Trigonal planar, 120° |
| Acetaldehyde, CH₃CHO | the carbonyl carbon | 3 + 0 = 3 | sp² | Trigonal planar |
| Acetaldehyde, CH₃CHO | the methyl carbon | 4 + 0 = 4 | sp³ | Tetrahedral |
| Methyl carbocation, CH₃⁺ | the carbon | 3 + 0 = 3 | sp² | Trigonal planar, flat |
| Methyl carbanion, CH₃⁻ | the carbon | 3 + 1 = 4 | sp³ | Trigonal pyramidal |
| Allene, CH₂=C=CH₂ | central carbon | 2 + 0 = 2 | sp | Linear |
| Allene, CH₂=C=CH₂ | each end carbon | 3 + 0 = 3 | sp² | Trigonal planar |
Two rows deserve a second look. The carbocation and carbanion have the same skeleton, but the carbanion's lone pair pushes the steric number from 3 to 4 — so it is sp³ and pyramidal instead of sp² and flat. And allene has three carbons in a row with two different hybridisations, which forces its two CH₂ groups into perpendicular planes. Hybridisation belongs to an atom, not to a whole molecule.
What s character actually predicts
An s orbital sits closer to the nucleus than a p orbital, so the more s character a hybrid orbital has, the more tightly its electrons are held. Three examinable trends follow:
| Bond | Hybridisation | C–C bond length | C–H bond length |
|---|---|---|---|
| C–C in ethane | sp³–sp³ | 154 pm | 109 pm |
| C=C in ethene | sp²–sp² | 134 pm | 108 pm |
| C≡C in ethyne | sp–sp | 120 pm | 106 pm |
1. Bonds get shorter and stronger as s character rises — partly because the extra π bonds pull the nuclei together, partly because the sigma bond itself is tighter.
2. Effective electronegativity rises in the order sp³ < sp² < sp.
3. Terminal alkynes are acidic. Approximate pKa values are about 50 for ethane, 44 for ethene and 25 for ethyne. The sp carbanion holds its lone pair in an orbital with 50% s character, close to the nucleus and therefore stabilised — which is why ethyne reacts with sodium metal to give sodium acetylide, and ethane does not.
Same element, two famous solids
In diamond every carbon is sp³ and bonded to four others in a rigid three-dimensional network. There are no free electrons, so diamond does not conduct electricity, and the network makes it extremely hard. Graphite is built from flat sheets of sp² carbons in hexagons; each keeps one unhybridised p orbital, and those overlap sideways across the whole sheet to give delocalised electrons that conduct. Weak forces between the sheets let them slide, which is why graphite is a lubricant and a pencil works. One element, two hybridisation states, completely different properties.
Common mistakes that cost marks
- Counting pi bonds in the steric number. A C=C carbon has a steric number of 3, not 4. Only sigma bonds and lone pairs count.
- Forgetting lone pairs. They occupy hybrid orbitals just like bonds. Missing the lone pair on a carbanion turns a correct sp³ into a wrong sp².
- Assigning one hybridisation to a whole molecule. In CH₃CHO one carbon is sp³ and the other is sp². Always say which carbon.
- Quoting the ideal angle when lone pairs are present. 109.5°, 120° and 180° are the ideal values for the pure cases; lone pairs compress them.
- Writing sp3 as "s + p + 3". The superscript is the number of p orbitals used: sp³ means one s and three p, giving four orbitals in total.
- Saying sp² carbon has three bonds. It has three sigma bonds plus one pi bond — four bonds altogether, because carbon is always tetravalent in neutral molecules.
Where hybridisation appears in exams
| Exam | Typical question |
|---|---|
| CBSE / ICSE Class 11 | State the hybridisation and shape of a given carbon; explain the structure of ethene and ethyne |
| CBSE / ICSE Class 12 | Hybridisation in carbonyl compounds, benzene and reaction intermediates |
| JEE / NEET | Assign hybridisation across a whole molecule; bond length and acidity comparisons |
| IIT-JAM / CUET-PG | Carbocation and carbanion geometry, allene stereochemistry, conjugation |
| GATE / CSIR-NET | Orbital overlap arguments in mechanism, aromaticity and spectroscopic correlations |
Start from the electron configuration. The Interactive Periodic Table shows carbon's 1s² 2s² 2p² configuration, its electronegativity and its atomic radius — the three facts the whole hybridisation argument is built on — and lets you compare carbon with nitrogen, oxygen and silicon in the same view.
Open the Interactive Periodic Table →Want bonding and organic chemistry taught from first principles? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in Delhi-NCR — details at abcchemistry.in.